(a) `A` solution is prepared by dissolving `10g` of non-volatile solute in `180g` of `H_(2)O`. If the relative lowering of vapour pressure is `0.005`,

Question

(a) `A` solution is prepared by dissolving `10g` of non-volatile solute in `180g` of `H_(2)O`. If the relative lowering of vapour pressure is `0.005`,

(a) `A` solution is prepared by dissolving `10g` of non-volatile solute in `180g` of `H_(2)O`. If the relative lowering of vapour pressure is `0.005`, find the molar mass of the solute.
(b) How many grams of sucrose `(C_(12)H_(22)O_(11))`must be dissolved in `90g` of water to produce a solution over which the relative humidity is `80%`? Assume the solution is ideal.

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

Related Questions

The vapour pressure of pure benzeen at `50^(@)C` is 268 torr. How many mol of non-volatile solute per mol of benzene is required to prepare a solution

2 Answers 5 Views

If `P^(@)` the vapour pressure of a pure solvent and P is the vapour pressure of the solution prepared by dissolving a non volatile solute in it. The

2 Answers 4 Views

An ideal solution was prepared by dissolving some amount of cane sugar (non-volatile) in `0.9` moles of water. The solution was then cooled just below

2 Answers 4 Views

Calculate the weight of non`-` volatile solute having molecular weight 40, which should be dissolvd in `57 gm` octane to reduce its vapour pressure to

2 Answers 4 Views

The mass of a non-volatile solute (Molar mass 40) which should be dissolved in `114g` octance to reduce its vapour pressure to `80%` is `2xg`. Find th

2 Answers 7 Views

How is relative lowering of vapour pressure defined for a solution consisting of volatile solvent and non-volatile solute ? How is this function relat

2 Answers 5 Views

Relative decrease in vapour pressure of an aqueous `NaCl` is `0.167`. Number of moles of `NaCl` present in `180g` of `H_(2)O` is:

2 Answers 6 Views

The vapour pressure of pure benzene, `C_(6)H_(6)` at `50^(@)C` is 268` Torr. How many moles of non-volatile solute per mol of benzene is required to p

2 Answers 4 Views

The vapour pressure of a solution of a non-volatile electrolyte `B` in a solvent `A` is `95%` of the vapour pressure of the solvent at the same temper

2 Answers 4 Views

The relative lowering of vapour pressure of an aqueous solution containing a non-volatile solute, is 0.0125. The molality of the solution is

2 Answers 4 Views

Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - (a) `199 g//"mole"`, (b) `427.5gm`
(a) Relative lowering of vapour pressure `=X_(A)`
or `0.005=(10//M_(A))/(10//M_(A)+180//18)` (Let `M_(A)` is the molar mass of solute)
so `M_(A)=199 gm//"mole"`
(b) Let `P^(@)` be the vapour pressure of water, now if relative humidity is `80%` after dissolving `C_(12)H_(22)O_(11)`. then the new pressure of solution will be `P` (let).
so `(P)/(P_(0))=0.8` or `P=0.8P_(0)`
Now Relative lowring of pressure `=(P_(0)-P)/(P_(0))=X_(A)`
so `(P_(0)-0.8P_(0))/(P_(0))=(x)/(342/(x/(342)+(90)/(18)))` {let `x` be the `wt.` of `C_(12)H_(22)O_(11)` dissolved}
`therefore x=427.5gm`.